Prove that the divergence of a curl is always zero. Checkit for function ${\mathbf{V}}_{\mathbf{a}}$ in Prob. 1.15.

The divergence of curl of a unction is defined as$\nabla .(\nabla \times v)$ . The vector is defined as $v=v_{x}i+v_{y}j+v_{z}k.$ the $\nabla$operator is defined as

$\nabla =\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k.$

The curl of vector $\overrightarrow{v}$is computed as follows:

$$\begin{gathered} \nabla xv=\left|\begin{array}{ccc}i& j& y\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ v_x& v_y& v_z\end{array}\right| \\ =\left(\frac{\partial v_z}{\partial y}-\frac{\partial v_y}{\partial z}\right)i-\left(\frac{\partial v_z}{\partial x}-\frac{\partial v_x}{\partial z}\right)j+\left(\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}\right)k \end{gathered}$$

Now compute divergence of curl of vector $\overrightarrow{v}$,as :

$$\begin{gathered} \nabla .(\nabla \times v)=\left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}+j\frac{\partial }{\partial z}k\right).\left(\left(\frac{\partial v_z}{\partial y}-\frac{\partial v_y}{\partial z}\right)i-\left(\frac{\partial v_z}{\partial x}-\frac{\partial v_x}{\partial z}\right)j+\left(\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}\right)k\right) \\ =\frac{\partial }{\partial x}\left(\frac{\partial v_z}{\partial y}-\frac{\partial v_y}{\partial z}\right)+\frac{\partial }{\partial y}\left(\frac{\partial v_z}{\partial x}-\frac{\partial v_x}{\partial z}\right)+\frac{\partial }{\partial z}\left(\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}\right) \\ =0 \end{gathered}$$

Thus the value of divergence of cutl of a function is 0.

To compute an expression substitute the vectors and other required expression and then simplify.

The vector $\overrightarrow{v}$ is defined as $x^{2}i+3x{z}^{2}j-2xzk$. The ldivergence of curl of the vector $\overrightarrow{v}$is computed as follows:

$$\begin{gathered} \nabla .(\nabla \times v_a)=\frac{\partial }{\partial x}\left(\frac{\partial v_z}{\partial y}-\frac{\partial v_y}{\partial z}\right)+\frac{\partial }{\partial y}\left(\frac{\partial v_z}{\partial x}-\frac{\partial v_x}{\partial z}\right)+\frac{\partial }{\partial z}\left(\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}\right) \\ =\frac{\partial }{\partial x}\left(\frac{\partial (-2xz)}{\partial y}-\frac{\partial \left(3x{z}^2\right)}{\partial z}\right)+\frac{\partial }{\partial y}\left(\frac{\partial (-2xz)}{\partial x}-\frac{\partial \left(x^2\right)}{\partial z}\right)+\frac{\partial }{\partial z}\left(\frac{\partial \left(3x{z}^2\right)}{\partial x}-\frac{\partial \left(x^2\right)}{\partial y}\right) \\ =\frac{\partial }{\partial x}(0-6xz)+\frac{\partial }{\partial y}(0-(-2z\left)\right)+\frac{\partial }{\partial z}(3z^2-0) \\ =-6z-0+6z \\ =0 \end{gathered}$$

Thus the value of $\nabla .(\nabla \times v_a)$is 0.